Question

我一直在努力获取与Iron一起使用的持久性参考，不确定如何设置适当的寿命。我希望能够在不同的路线上重复使用控制器。

示例：

use iron::prelude::*;
use iron::typemap::Key;
use persistent::Read;
use router::Router;

pub struct Controller;

pub struct Rest {
    controller: Controller,
}

impl Key for &Controller {
    type Value = Self;
}

impl Rest {
    pub fn run(&self) {
        let router = Router::new();
        let mut chain = Chain::new(router);
        chain.link(Read::<&Controller>::both(&self.controller));
        Iron::new(chain).http(format!("0.0.0.0:1234")).ok();
    }
}

fn main() {
    Rest {
        controller: Controller,
    }
    .run();
}

[dependencies]
iron = "0.6.*"
router = "0.6.*"
persistent = "0.4.0"

Gist of example

error[E0478]: lifetime bound not satisfied
  --> src/main.rs:12:6
   |
12 | impl Key for &Controller {
   |      ^^^
   |
note: lifetime parameter instantiated with the lifetime '_ as defined on the impl at 12:14
  --> src/main.rs:12:14
   |
12 | impl Key for &Controller {
   |              ^
   = note: but lifetime parameter must outlive the static lifetime

error[E0495]: cannot infer an appropriate lifetime for lifetime parameter `'_` due to conflicting requirements
  --> src/main.rs:12:6
   |
12 | impl Key for &Controller {
   |      ^^^
   |
note: first, the lifetime cannot outlive the lifetime '_ as defined on the impl at 12:14...
  --> src/main.rs:12:14
   |
12 | impl Key for &Controller {
   |              ^
   = note: ...so that the types are compatible:
           expected typemap::Key
              found typemap::Key
   = note: but, the lifetime must be valid for the static lifetime...
note: ...so that the type `&Controller` will meet its required lifetime bounds
  --> src/main.rs:12:6
   |
12 | impl Key for &Controller {
   |      ^^^

Answer 1

错误消息指出：

但是有效期参数必须超过静态有效期

这是因为Key使用Any作为超级特征，这需要'static：

pub trait Any: 'static {
    fn type_id(&self) -> TypeId;
}

最简单的解决方案是为一个值实现Key，然后将该值提供给Read::both：

impl Key for Controller {
    type Value = Self;
}

impl Rest {
    pub fn run(self) {
        let router = Router::new();
        let mut chain = Chain::new(router);
        chain.link(Read::<Controller>::both(self.controller));
        Iron::new(chain).http(format!("0.0.0.0:1234")).ok();
    }
}

我希望在所有这些路由中全局共享持久数据

在那种情况下，我会完全避免使用永久性板条箱，而只是创建一个单例：

Answer 2

您需要为impl块添加一个明确的生命周期绑定。否则，编译器不了解控制器参考的参考有效性。

这应该有效，尽管我尚未测试

pub struct Rest {
    controller: controller::Controller,
}

impl<'a> Key for &'a controller::Controller {
    type Value = Self;
}

impl Rest {
    pub fn run(&self) {
        let mut router = Router::new();
        let mut chain = Chain::new(router);
        chain.link(Read::<&controller::Controller>::both(&self.controller));
        Iron::new(chain).http(format!("0.0.0.0:1234"))
    }
}

另一方面，我认为您不需要在Read调用中指定类型，编译器应该能够从上下文中确定它。

具有结构引用的铁持久状态

2 个答案: